I have the following integral \begin{align*} \int_{-\infty}^\infty f(t) q(t+ax) dt \end{align*}

where a is some constant.

This integral look a lot like convolution (or correlation). My question is can it be re-write as an convolution of $ f(???)* g(???)$? What are the arguments of the two functions? For example if the integral was $\int_{-\infty}^\infty f(t) q(at+ax) dt$ then we can re-write as \begin{align*} \int_{-\infty}^\infty f(t) q(at+ax) dt= f(x)*q(ax) \end{align*}

This is what I tried I did some integral manipulation and I got this identity \begin{align*} \int f(t) q(t+ax) dt=\int f(t-ax) q(t) dt \end{align*}


A convolution integral is of the form

$$\int f(t) g(x-t) \, dt;$$

the important point is the structure of the arguments:

$$x-t \qquad \text{and} \qquad t$$

In order to write the given integral as a convolution integral, we define $\tilde{q}(t) := q(-t)$. Then

$$\int_{-\infty}^{\infty} f(t) q(t+ax) \, dt = \int_{-\infty}^{\infty} f(t) \tilde{q}(-ax-t) \, dt = (f \ast \tilde{q})(-ax).$$

  • $\begingroup$ So is this a convolution of $f(x)$ with $\bar{q}(-ax)=q(ax)$??? $\endgroup$ – Boby Dec 28 '14 at 15:47
  • $\begingroup$ @Boby Note that $(f*\tilde{q})(-ax) \neq (f*q)(ax)$. The first expression means: Calculate the convolution of $f$ with $\tilde{q}$ and then evaluate this function at $-ax$. $\endgroup$ – saz Dec 28 '14 at 16:13
  • $\begingroup$ You see I want apply fourier transform and use the fact that convolution is a product. I have to know what the time domain functions $\endgroup$ – Boby Dec 28 '14 at 16:15
  • $\begingroup$ @Boby So you have to calculate the Fourier transform of $\tilde{q}$. Where are you stuck? $\endgroup$ – saz Dec 28 '14 at 16:34
  • 1
    $\begingroup$ @Boby It is gonna be $\frac{1}{|a|}F(\frac{\omega}{-a})\tilde{Q}(\frac{\omega}{-a})$ $\endgroup$ – Sina Dec 29 '14 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.